draw a line perpendicular to and . Let be the point, where it cuts thus last, plane, the "base", and the point where this plane is encountered by the generating line through . If then , we have

(13) |

The strokes over the letters indicate the absolute values of the distances and .

It can be shown (§ 8) that, all quantities being expressed in natural units, the "volume" of the prism is found by taking the product of the numerical values of the base and the "height" .

Let now linear three-dimensional extensions perpendicular to be made to pass through and . From these extensions the lateral boundary of the prism cuts the parts and and these parts, together with the lateral surface, enclose a new prism , the volume of which is equal to that of . As now the volume of is given by the product of and , we have with regard to (13)

If now we remember that, if a vector perpendicular to is projected on the generating line, the ratio between the projection and the vector itself (viz. between their absolute values) is given by and that a connexion similar to that which was found above between a normal section of the prism and , also exists between and any other oblique section, we easily find the following theorem:

Let and be two arbitrarily chosen linear three-dimensional sections of the prism, and two vectors, perpendicular to and resp. and of the same length, and the absolute values of the projections of and on a generating line. Then we have

(14) |

§ 19. After these preliminaries we can show that the left hand side of (10) is equal to 0, if the numbers are constants and if moreover both the rotation and the rotation are everywhere the same. For the two parts of the integral the proof may be given in the same way, so that it suffices to consider the expression

(15) |

Let be the components of the vector , expressed in -units. From the distributive property of the vector product it then follows that each of the four components of